CODEWITHSUNDEEP
javascriptphpajaxdom
g770like
g770like

Reputation: 21

jquery not change input value from calling ajax page

Problem solved. Check working code on bottom.

first, sorry for my english. If possible pls rewrite my question.

I can not change input value with jquery from calling ajax page.

(jquery version http://ajax.googleapis.com/ajax/libs/jquery/1.3.0/jquery.min.js)

My main page is index.php and ajax page is ajax.php. I called ajax.php with

$.ajax(...

But after working ajax codes then input value didnt change on index.php.

here is index.php input code

<input name="guncel_fiyat" type="hidden" id="guncel_fiyat" value="old value">

here is index.php ajax calling code:

$.ajax(
            {
                type: "POST",
                url: "../ajax.php",
                data: post_edilecek_veri,
                cache: false,
                success: function(donen_veri)
                {
                    some code here;
                }
            }

here is ajax.php code (i did try 3 different method but nothing change)

$(".guncel_fiyat").val("new value");

or

$("#guncel_fiyat").val("new value");

or

$(".guncel_fiyat input").val("new value");

after calling ajax then try get new value on index.php but only getting "old value" val.

var guncel_fiyat_degeri = $("#guncel_fiyat").val();
alert(guncel_fiyat);

HERE IS WORKING CODE

CALL_AJAX.HTML:

   <!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.0/jquery.min.js"></script>
<script type="text/javascript">
function peyfunc() 
{ 
    "use strict";
    var current_price = $("#current_price").val();
    alert ("current_price="+current_price);
    $.ajax(
            {
                url: "ajax_test.php",
                cache: false,
                success: function(html)
                {
                    $("#display").html(html);
                }
            }   
    );

}
</script>
</head>
<body>
<form id="form1" name="form1" method="post">
  <input name="current_price" type="hidden" id="current_price" value="5">
  <input type="button" name="button" id="button" value="Button"  onclick="peyfunc()">
</form>
<div id="display"></div>
</body>
</html>

AJAX_TEST.PHP

<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.0/jquery.min.js"></script>
test
<script type="text/javascript">
//alert("aaa");
$("#current_price").val("new value");
alert ($("#current_price").val());
</script>

Upvotes: 0

Views: 1707

Answers (1)

g770like
g770like

Reputation: 21

HERE IS WORKING CODE

CALL_AJAX.HTML:

   <!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.0/jquery.min.js"></script>
<script type="text/javascript">
function peyfunc() 
{ 
    "use strict";
    var current_price = $("#current_price").val();
    alert ("current_price="+current_price);
    $.ajax(
            {
                url: "ajax_test.php",
                cache: false,
                success: function(html)
                {
                    $("#display").html(html);
                }
            }   
    );

}
</script>
</head>
<body>
<form id="form1" name="form1" method="post">
  <input name="current_price" type="hidden" id="current_price" value="5">
  <input type="button" name="button" id="button" value="Button"  onclick="peyfunc()">
</form>
<div id="display"></div>
</body>
</html>

AJAX_TEST.PHP

<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.0/jquery.min.js"></script>
test
<script type="text/javascript">
//alert("aaa");
$("#current_price").val("new value");
alert ($("#current_price").val());
</script>

Upvotes: 1

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